Group theory

Question

Let G be any abelian group and π : G → G a group homomorphism satisfying π2 = π
(i.e. π(π(x) = π(x) for all x ε G). Show that G is the internal direct sum of ker π and Im π.

Answer 1

Take x ∈ G. Note π(x - π(x)) = π(x) - π(π(x)) = π(x) - π(x) = 0, so that x - π(x) ∈ ker π. Now x = π(x) + (x - π(x)), so x ∈ Im π + ker π. Hence G = Im π + ker π. Further, if y ∈ ker π ∩ Im π we have π(y) = 0 and y = π(z) for some z ∈ G. Therefore, y = π(z) = π(π(z)) = π(y) = 0 and we deduce ker π ∩ Im π = 0. This means that G = ker π ⊕ Im π, as desired.